If we look at the inside sum we have $$ \sum_{k=1}^nk\cdot\sum_{i=1}^{n-k+1}i $$ Now we have that the inside sum is the triangle numbers namely we have that $$ \sum_{i=1}^{n-k+1}i=\frac{(n-k+1)(n-k+2)}{2} $$ Thus, we have that the sum becomes $$ \sum_{k=1}^n\frac{k(n-k+1)(n-k+2)}{2} $$ Now if we factor everything out and gather things in terms of $k$ we get $$ \frac{1}{2}\sum_{k=1}^nk^3-(2n-3)k^2+(3n+2)k $$ Then if we break up the sums we have $$ \frac{1}{2}\sum_{k=1}^nk^3 -\frac{2n-3}{2}\sum_{k=1}^nk^2+\frac{3n+2}{2}\sum_{k=1}^nk $$ Now if we use the well known formulas $$ \sum_{k=1}^nk^3=\left(\frac{n(n+1)}{2}\right)^2 $$

$$ \sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}{6} $$ and the formula for the triangle numbers above, we then see that the sum becomes $$ \frac{1}{2}\left(\frac{n(n+1)}{2}\right)^2-\frac{2n-3}{2}\frac{n(n+1)(2n+1)}{6}+\frac{3n+2}{2}\frac{n(n+1)}{2} $$ Now we are really only interested in the $n^4$ term as the lower order terms will be killed when we take the limit.

Thus, we have the above is $$ -\frac{1}{24}n^4+O(n^3) $$ Thus, when we consider $$ \lim_{n\rightarrow \infty}\frac{96}{n^4}\cdot\text{sum}=\lim_{n\rightarrow \infty}\frac{96}{n^4}\cdot\left(-\frac{1}{24}n^4+O(n^3)\right)=-4 $$

If we look at the inside sum we have $$ \sum_{k=1}^nk\cdot\sum_{i=1}^{n-k+1}i $$ Now we have that the inside sum is the triangle numbers namely we have that $$ \sum_{i=1}^{n-k+1}i=\frac{(n-k+1)(n-k+2)}{2} $$ Thus, we have that the sum becomes $$ \sum_{k=1}^n\frac{k(n-k+1)(n-k+2)}{2} $$ Now if we factor everything out and gather things in terms of $k$ we get $$ \frac{1}{2}\sum_{k=1}^nk^3-(2n+3)k^2+(n^2+3n+2)k $$ Then if we break up the sums we have $$ \frac{1}{2}\sum_{k=1}^nk^3 -\frac{2n+3}{2}\sum_{k=1}^nk^2+\frac{n^2+3n+2}{2}\sum_{k=1}^nk $$ Now if we use the well-known formulas $$ \sum_{k=1}^nk^3=\left(\frac{n(n+1)}{2}\right)^2 $$

$$ \sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}{6} $$ and the formula for the triangle numbers above, we then see that the sum becomes $$ \frac{1}{2}\left(\frac{n(n+1)}{2}\right)^2-\frac{2n+3}{2}\frac{n(n+1)(2n+1)}{6}+\frac{n^2+3n+2}{2}\frac{n(n+1)}{2} $$ Now we are really only interested in the $n^4$ term as the lower order terms will be killed when we take the limit.

If we see from the first term we will get a factor of $\frac{n^4}{8}+O(n^3)$, the second factor, we will get a factor of $-\frac{n^4}{3}+O(n^3)$, and the last term will give us a $\frac{n^4}{4}+O(n^3)$

Thus, we have the above is $$ \frac{1}{24}n^4+O(n^3) $$ Thus, when we consider $$ \lim_{n\rightarrow \infty}\frac{96}{n^4}\cdot\text{sum}=\lim_{n\rightarrow \infty}\frac{96}{n^4}\cdot\left(\frac{1}{24}n^4+O(n^3)\right)=4 $$